Question 294874
Ugh. Please use more parentheses, they're free and help avoid confusion. 
I think this is what you want. If not please re-post.
{{{7^(-x-3)*7^(1+x)-4=0}}}
{{{7^(-x-3+1+x)-4=0}}}
{{{7^(-2)-4=0}}}
{{{1/49=4}}}
Clearly this is not true. There is no solution. 
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<<----- ADDITIONAL INFORMATION ----->>

>> This is what I mean.  {{{7^(-x)-3*7^(1+x)-4=0 }}}<<

OK, that's clearer. 
{{{ 7^(-x)-3*7^(1+x)-4=0}}}
Again, multiply by {{{ 7^(x)}}}.

{{{ 1-3*7^(2x+1)-4*7^(x)=0}}}

Pull out a 7 from the second term.

{{{ 1-3*7*7^(2x)-4*7(x)=0}}}

Re-arrange.

{{{ -21*7^(2x)-4*7^(x)+1=0}}}

{{{ 21*7^(2x)+4*7^(x)-1=0}}}

Substitute, let {{{ u=7^(x)}}}, then {{{ u^2=7^(2x)}}}

{{{ 21u^2+4u-1=0}}}

This quadratic is factorable.

{{{ 21u^2+4u-1=(7u-1)(3u+1)=0}}}

Two solutions using the zero product rule:

{{{ 7u-1=0}}}

{{{ u=1/7}}}

Now substitute back

{{{ 7^(x)=1/7}}}

{{{ x=-1}}}

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{{{ 3u+1=0}}}

{{{ u=-1/3}}}

{{{ 7^(x)=-1/3}}}

No solution here, since log requires positive arguments.

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Verify the one solution.

{{{ x=-1}}}

{{{ 7^(-x)-3*7^(1+x)-4=0}}}
{{{ 7^(1)-3*7^(0)-4=0}}}

{{{ 7-3-4=0}}}

{{{ 0=0}}}

Good solution.

{{{ highlight(x=-1) }}}